KAD 3131 : Laboratory 2

Structure Laboratory

1.0 INTRODUCTION

The two-hinged arch is a more efficient structure than the corresponding three
hinged arch. It is statically indeterminate by one degree; the horizontal reaction at the
hinged which are assumed to be fixed in position is the external force which cannot be
derived from the equations of equilibrium.
The typical solution is to form the expression for the horizontal displacement
at a hinge and so equate this to zero. The bending moment is the derived by
superimposing that due to the horizontal reaction on the diagram obtained from applied
loading acting on a simply supported beam with the same span as the arch. The
mathematical formula is: -
M =M ss −H . y

Where y = rise of the arch above the hinges. It will be seen that to a suitable scale H.y is
the arch outline.
Of the three traditional forms of an arch namely semi-circular, parabolic and semi-
elliptical, the first two can be analyzed quite simply. Generally, the choice between them
is determined by the difference in levels of terrain surrounding the bridge. It would be
more probable to find a semi-circular arch crossing a ravine where the depth of the
structure was a free choice for the designer. A further factor is the reduction in horizontal
restraint as the rise/ span ratio increases.
The experiment arch affords the opportunity to test how a difference between the
model and its analysis introduces errors. The simplified analysis assumes that the arch
cross section varies to keep ds/l constant, whereas the model has a uniform cross
section.

2.0 OBJECTIVE

To compare the experimental values of horizontal reaction with the simplified theory,
and tp provide the influence line for this reaction.

3.0 THEORY

The two hinged arch is a statically indeterminate structure of the first degree. The
horizontal thrust is the redundant reaction and is obtained y the use of strain energy
methods.

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 KAD 3131 : Laboratory 2
Structure Laboratory

B
A B
L

1
Fig (a)

H A B H

B 2

Fig (b)
Two hinged arch is made determinate by treating it as a simply supported curved
beam and horizontal thrust as a redundant reaction. The arch spreads out under external
load as shown in fig. (a). This results in a horizontal displacement of support B by ∆1.
Here, deflection due to flexure only has been considered. Since the support conditions
dictate that that the final displacement at support B should be zero, horizontal reaction
H should be such that displacement ∆2 caused by H must satisfy the condition.
∴ ∆1 +∆ 2=0

∆1 + f × H=0

Where, f is the displacement caused by a unit force applied in the direction of H.

−∆ 1
∴H= (1)
f

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 KAD 3131 : Laboratory 2
Structure Laboratory

Therefore, it is required to calculate the horizontal displacement in arch caused by

external load as well as unit horizontal force.
The horizontal displacement in a curved member can be found by either Castiglione’s
second theorem or the unit load method.
B
δM ds
∴ H =∫ M
A δH EI
B
ds
¿∫ Mm
A EI

B
ds
similarlu f =∫ m2
A EI

Mm ds
EI∫
therfore , H= 2
(2)
m ds
∫ EI
Where,
M = Bending moment on any point on the arch due to given loading.
m = Moment on any point on the arch due to a unit horizontal force
applied
at B in the direction of H.
The expression given by Eq. (2) will become simpler provided the curve of the arch
axis is parabolic and moment of inertial of curve at any section varies as I=I_0secθ where
θ is angle between the horizontal and tangent to the arch axis at that particular point.
IO= moment of inertia at the crown
I = moment of inertia at any other section
m= y, ds = sec θ dx
it may be noted that the integration is to be carried out from 0 to L then Eq. (2) will
become
L

∫ my dx
H= OL (3)
∫ y 2 dx
O

L
my dx
and ∆1 = horizontal displacement ¿∫
O E I0

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 KAD 3131 : Laboratory 2
Structure Laboratory

For a concentrated load, W at the crown it is found that

2
5 WLr
Horizontal displacement , ∆ 1= (4)
48 E I 0

Where, L is the span of the arch and r is the rise.

The horizontal movement of the roller end can be found by this method for any
position of the load on the arch.

Fig (c)
The ordinate for the influence line diagram for H at any distance z = aL form L.H.S.
can be obtained as follows:
Wx
M= ( L−z ) for 0 < x< z
L
Wz
M= ( L−x ) for 0< x< L
L
Now H can be evaluated using Eq. (3)
L Z L

∫ My dx=∫ Wx
O O L
x x2
[
( L−z ) 4 r − 2 dx +∫
L L Z
]
Wz ( L−x )
L
x x2
4 r − 2 dx
L L [ ]
Wz ( L−z)(L2 + Lz−Z 2)
¿
3 L2

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 KAD 3131 : Laboratory 2
Structure Laboratory

[ ( )]
L L 2
2 x x
∫y dx=∫ 4r − 2
L L
dx
O O

8
¿ Lr 2
15

L2 +Lz−Z2
−5 WL
∴H= (L−z )¿
8 r
Now substituting z=aL

5 WL
H=− (a−2a 3 +a 4 )
We have 8 r

Taking W = 1kg
Influence line ordinate are given by
5 L
ILO = (a−2 a3 + a4 )
8 r

4.0 APPARATUS

Figure 1
 1 HSC5m/f Dial Gauge Assembly
 1 HST501P Arch with Side Bracket
 1 HST502 Track Plate Assembly with Dial Gauge
 1 HWH5 Reaction Load Hanger 2N
 7 HWH4 Load Hanger 1N

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 KAD 3131 : Laboratory 2
Structure Laboratory

 1 HST405W Set of Weights

5.0 METHODOLOGY

Part 1: Horizontal Reaction for Point Load at 3/8 Span

1. With a load hanger on one of the stirrups near the middle of the arch, the dial gauge
datum reading at the moving hinged was noted.
2. 10N increments of vertical load were applied to the hanger up to a total not
exceeding 50N.
3. As each increment was added, loads were applied to the horizontal reaction – hanger
to restore the dial reading to the datum value, thus complying theoretical condition
of zero displacement of the hinges.
4. The horizontal reactions were recorded in Table 1.
Part 2: Theoretical Reaction for a Uniform Load
1. With load hangers on all seven stirrups the dial gauge datum was noted.
2. A uniformly distributed load was stimulated by placing 5N on each hanger.
3. The horizontal reaction was found to give zero displacement of the moving hinge.
4. The loading was increased to 10N per hanger and again the hinge was restored to its
datum position.
5. The results were recorded in Table 2.
Part 3: Influence Line for H
1. With all seven load hangers still in the stirrups, the dial gauge datum was noted.
2. The horizontal reaction was found for zero displacement as a 50N load was placed on
each hanger in turn from left to right.
3. The results were recorded in Table 3.

6.0 DATA AND RESULTS

Point Load (N) Horizontal Reaction Theoretical Reaction Percentage of

(N) (N) Error (%)
0 0.0 0.0 0.00
10 8.5 9.4 9.57
20 17.1 18.8 9.04
30 25.7 28.1 8.54
40 34.3 37.5 8.53
50 43.2 46.9 7.89

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 KAD 3131 : Laboratory 2
Structure Laboratory

Table 1: Horizontal Reaction for Point Load at 3/8 Span

Distributed Load Dial Gauge Horizontal Theoretical Percentage

(with hanger) Datum Reaction Reaction Error
(with hanger) (N) (N) (%)
7 ×5 N 1345 24 25 4.0
7 ×10 N 1345 48 50 4.0
Table 2: Horizontal Reaction for Uniformly Distributed Load

Position of 50N Load as Horizontal Reaction (N) Influence Line Ordinate

Fraction of Span (m)
0.000 0 0.00
0.125 18.5 0.37
0.250 33.8 0.68
0.375 43.4 0.87
0.500 47.2 0.94
0.625 43.4 0.87
0.750 33.8 0.68
0.875 18.5 0.37
1.000 0.0 0.00
Table 3: Influence Values of Horizontal Reaction

Horizontal Reaction against Point Load

50

45

40

35
Horizontal Reaction, N

30

25

20

15

10

0
0 10 20 30 40 50 60

Point Load, N

Hori zontal Reaction (N) Theoretica l Rea ction (N)

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 KAD 3131 : Laboratory 2
Structure Laboratory

Influence Line Ordinate against Horizontal Reaction

1
0.9
0.8
0.7
Influence Line

0.6
0.5
0.4
0.3
0.2
0.1
0
0 18.5 33.8 43.4 47.2 43.4 33.8 18.5 0
Horizontal Reaction, N

Influence Li ne Ordi nate

Sample Calculation
Horizontal Reaction for Point Load at 3/8 Span
Experimental
y 2− y 1
Gradient=
x 2−x 1
34.5−15
Gradient=
40−17
¿ 0.848

Theoretical

∑ M =0 , ∑ F y =0 , ∑ M C =0

Load = 10 N

∑ M A =0
R
(¿¿ B) ( 1.000 )−(10)(0.625)
0=¿
RB =6.25 N

∑ F y =0
0=R A−10+ 6.25

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 KAD 3131 : Laboratory 2
Structure Laboratory

R A =3.75 N

∑ M C=0
6.25
)(0.500)
0=( HR ) ( 0.2 ) + ( 10 ) ( 0.125 )−¿
HR=9.4 N

Percentage of error=| Experimental−Theory

Theory |× 100
Percentage of error=|
9.4 |
8.5−9.4
×100

¿ 9.57

Horizontal Reaction for Uniformly Distributed Load (Theoretical)

∑ M =0 , ∑ F y=0 , ∑ M C=0
Load 5N

∑ M A =0
R
(¿¿ B) 1.000 − 5 0.125 − 5 0.250 − 5 0.375 )−( 5 ) ( 0.5 .00 )−( 5 )( 0.625 )−( 5 ) ( 0.750 )−( 5 )( 0.875 )
( ) ( )( ) ( ) ( ) ( )(
0=¿
RB =17.5 N

∑ F y =0
0=R A−5−5−5−5−5−5−5+17.5

R A =17.5 N

∑ M C=0
17.5
)(0.5)
0=( HR ) ( 0.2 )− (5 )( 0.125 )−( 5 ) ( 0.25 )− (5 )( 0.375 )+ ¿
HR=−25.0 N =25.0 N

Percentage of error= |25.0−24.0

25.0 |
× 100=4.0

Influence Values of Horizontal Reaction

H R×2
Influence LineOrdinate=
100

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 KAD 3131 : Laboratory 2
Structure Laboratory

18.5 ×2
Influence LineOrdinate=
100
7.0 DISCUSSION AND RECOMMENDATION
Based on the results of the experiment, the highest percentage error for the
horizontal reaction for point load at 3/8 span was 9.37% while the least was 7.89%. On the
other hand, for the horizontal reaction for the uniformly distributed load was 4% for both 5N
and 10N loads. The results showed small percentage errors, thus the experiment was
successful.
Although the results of the experiment are accepted, there were still possibilities of
error done during the experiment. A possibility of error is incorrect reading of dial gauge by
the observer. Also, there is a possibility that the loads were not well distributed at its
position. Besides that, since the apparatus used have been used for quite a long time, there
is a possibility that the apparatus did not function as well as its supposed to be. These
possible errors could more or less affect the results of the experiment.
In order to obtain more accurate results, it is recommended that we check the
condition of the apparatus whether it is good or not before the experiment. The condition of
the equipment is most important as it is directly involved with the results. Also, when
putting the load at the hangers, do it slowly and carefully to make sure that the position is
accurate and the beam does not shake. Besides that, to reduce the effect of parallax error, it
is recommended to repeat the same load and position a few times and get an average value.
8.0 CONCLUSION
In conclusion, from the results of Part 1 and 2, we can conclude that the theory of
the two hinged parabolic arch for the horizontal reactions for point load and uniformly
distributed load is valid. Although the possibility of errors were present, the results were still
acceptable.
The two-hinged parabolic arch is the statically indeterminate structure to one
degree. Usually, the horizontal reaction is treated as the redundant and is evaluated by the
method of least work. The two hinged parabolic arch is most often used to bridge long
spans. This type of arch are connected with pins at the base. Pinned connections at the base
enable it to rotate, allowing the structure to move freely and compensate for thermal
expansion and contraction due to surrounding temperature.

9.0 REFERENCE/APPENDIX
2-Hinged Arch. URL: https://www.scribd.com/document/26761629/2-Hinged-Arch
Two Hinged Arches. URL: http://nptel.ac.in/courses/105105109/pdf/m5l33.pdf
Structural & Materials Laboratory Manual, Kartini Kamaruddin.

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